Proof that $A+A^T$ is singular












2












$begingroup$



Let $A in mathbb R^{10,10}$, $x_{1},x_{2},...,x_{7}in mathbb R^{10}$ which are linearly independent vectors and $Ax_{1}=Ax_{2}=...=Ax_{7}$. Proof that $A+A^T$ is singular (not invertible)




I am trying to do this task in two different ways and none of them works.


Firstly I wanted to show that $$det (A+A^T) = 0 $$
but there is no nice formula for $det A + B $ so probably it is wrong way.

From the other hand I have that idea:


Assume that exists $B$ such that
$$ Bcdot(A+A^T) = I_n$$
and now I wanted to proof that $$ker A^T subset ker A$$ then I could easly fail that assumption. But unfortunately It (my sub-proof) seems to be false theorem so... have somebody any idea how to solve this task?






linear-algebra matrices determinant







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    you mean is not invertible ? since $det(B) = 0$ means $B$ is not invertible
    $endgroup$
    – Thinking
    Feb 1 at 17:19










  • $begingroup$
    Sorry - I mean irvertible
    $endgroup$
    – VirtualUser
    Feb 1 at 17:20










  • $begingroup$
    What is irvertible?
    $endgroup$
    – Math Lover
    Feb 1 at 17:21










  • $begingroup$
    @MathLover full rank (for square matrices). Strictly speaking, $M$ is invertible if there exists $M^{-1}$ such that $MM^{-1}=M^{-1}M = I$.
    $endgroup$
    – Surb
    Feb 1 at 17:22


















2












$begingroup$



Let $A in mathbb R^{10,10}$, $x_{1},x_{2},...,x_{7}in mathbb R^{10}$ which are linearly independent vectors and $Ax_{1}=Ax_{2}=...=Ax_{7}$. Proof that $A+A^T$ is singular (not invertible)




I am trying to do this task in two different ways and none of them works.


Firstly I wanted to show that $$det (A+A^T) = 0 $$
but there is no nice formula for $det A + B $ so probably it is wrong way.

From the other hand I have that idea:


Assume that exists $B$ such that
$$ Bcdot(A+A^T) = I_n$$
and now I wanted to proof that $$ker A^T subset ker A$$ then I could easly fail that assumption. But unfortunately It (my sub-proof) seems to be false theorem so... have somebody any idea how to solve this task?






linear-algebra matrices determinant







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    you mean is not invertible ? since $det(B) = 0$ means $B$ is not invertible
    $endgroup$
    – Thinking
    Feb 1 at 17:19










  • $begingroup$
    Sorry - I mean irvertible
    $endgroup$
    – VirtualUser
    Feb 1 at 17:20










  • $begingroup$
    What is irvertible?
    $endgroup$
    – Math Lover
    Feb 1 at 17:21










  • $begingroup$
    @MathLover full rank (for square matrices). Strictly speaking, $M$ is invertible if there exists $M^{-1}$ such that $MM^{-1}=M^{-1}M = I$.
    $endgroup$
    – Surb
    Feb 1 at 17:22
















2












2








2





$begingroup$



Let $A in mathbb R^{10,10}$, $x_{1},x_{2},...,x_{7}in mathbb R^{10}$ which are linearly independent vectors and $Ax_{1}=Ax_{2}=...=Ax_{7}$. Proof that $A+A^T$ is singular (not invertible)




I am trying to do this task in two different ways and none of them works.


Firstly I wanted to show that $$det (A+A^T) = 0 $$
but there is no nice formula for $det A + B $ so probably it is wrong way.

From the other hand I have that idea:


Assume that exists $B$ such that
$$ Bcdot(A+A^T) = I_n$$
and now I wanted to proof that $$ker A^T subset ker A$$ then I could easly fail that assumption. But unfortunately It (my sub-proof) seems to be false theorem so... have somebody any idea how to solve this task?






linear-algebra matrices determinant







share|cite|improve this question











$endgroup$





Let $A in mathbb R^{10,10}$, $x_{1},x_{2},...,x_{7}in mathbb R^{10}$ which are linearly independent vectors and $Ax_{1}=Ax_{2}=...=Ax_{7}$. Proof that $A+A^T$ is singular (not invertible)




I am trying to do this task in two different ways and none of them works.


Firstly I wanted to show that $$det (A+A^T) = 0 $$
but there is no nice formula for $det A + B $ so probably it is wrong way.

From the other hand I have that idea:


Assume that exists $B$ such that
$$ Bcdot(A+A^T) = I_n$$
and now I wanted to proof that $$ker A^T subset ker A$$ then I could easly fail that assumption. But unfortunately It (my sub-proof) seems to be false theorem so... have somebody any idea how to solve this task?






linear-algebra matrices determinant




linear-algebra matrices determinant






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 1 at 17:25









Gabriel Romon

18k53387




18k53387










asked Feb 1 at 17:16









VirtualUserVirtualUser

1,316317




1,316317








  • 1




    $begingroup$
    you mean is not invertible ? since $det(B) = 0$ means $B$ is not invertible
    $endgroup$
    – Thinking
    Feb 1 at 17:19










  • $begingroup$
    Sorry - I mean irvertible
    $endgroup$
    – VirtualUser
    Feb 1 at 17:20










  • $begingroup$
    What is irvertible?
    $endgroup$
    – Math Lover
    Feb 1 at 17:21










  • $begingroup$
    @MathLover full rank (for square matrices). Strictly speaking, $M$ is invertible if there exists $M^{-1}$ such that $MM^{-1}=M^{-1}M = I$.
    $endgroup$
    – Surb
    Feb 1 at 17:22
















  • 1




    $begingroup$
    you mean is not invertible ? since $det(B) = 0$ means $B$ is not invertible
    $endgroup$
    – Thinking
    Feb 1 at 17:19










  • $begingroup$
    Sorry - I mean irvertible
    $endgroup$
    – VirtualUser
    Feb 1 at 17:20










  • $begingroup$
    What is irvertible?
    $endgroup$
    – Math Lover
    Feb 1 at 17:21










  • $begingroup$
    @MathLover full rank (for square matrices). Strictly speaking, $M$ is invertible if there exists $M^{-1}$ such that $MM^{-1}=M^{-1}M = I$.
    $endgroup$
    – Surb
    Feb 1 at 17:22










1




1




$begingroup$
you mean is not invertible ? since $det(B) = 0$ means $B$ is not invertible
$endgroup$
– Thinking
Feb 1 at 17:19




$begingroup$
you mean is not invertible ? since $det(B) = 0$ means $B$ is not invertible
$endgroup$
– Thinking
Feb 1 at 17:19












$begingroup$
Sorry - I mean irvertible
$endgroup$
– VirtualUser
Feb 1 at 17:20




$begingroup$
Sorry - I mean irvertible
$endgroup$
– VirtualUser
Feb 1 at 17:20












$begingroup$
What is irvertible?
$endgroup$
– Math Lover
Feb 1 at 17:21




$begingroup$
What is irvertible?
$endgroup$
– Math Lover
Feb 1 at 17:21












$begingroup$
@MathLover full rank (for square matrices). Strictly speaking, $M$ is invertible if there exists $M^{-1}$ such that $MM^{-1}=M^{-1}M = I$.
$endgroup$
– Surb
Feb 1 at 17:22






$begingroup$
@MathLover full rank (for square matrices). Strictly speaking, $M$ is invertible if there exists $M^{-1}$ such that $MM^{-1}=M^{-1}M = I$.
$endgroup$
– Surb
Feb 1 at 17:22












1 Answer
1






active

oldest

votes


















4












$begingroup$

In fact, the conclusion should be that $A+A^T$ is not invertible. Note that the range of $A$ is spanned by
$$
langle Ax_1,Ax_2,ldots , Ax_{10}rangle =langle Ax_7,Ax_8,Ax_{9}, Ax_{10}rangle,
$$ which implies that the rank of $A$ is at most $4$. By the rank theorem, $A^T$ has the same rank as $A$. This implies that $A+A^T$ has rank at most $text{rk}(A)+text{rk}(A^T)le 8<10$, which says that $A+A^T$ is not invertible.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    that's a nice proof!
    $endgroup$
    – Surb
    Feb 1 at 17:23










  • $begingroup$
    @Surb Thank you :-)
    $endgroup$
    – Song
    Feb 1 at 17:24






  • 1




    $begingroup$
    Made all the nicer by the fact its strategy also makes obvious why the original problem equated seven vectors; with just six, this proof wouldn't have worked.
    $endgroup$
    – J.G.
    Feb 1 at 17:28










  • $begingroup$
    So if $ker A$ has only $vec{0}$ then $A$ is irvertible?
    $endgroup$
    – VirtualUser
    Feb 1 at 17:29






  • 1




    $begingroup$
    @VirtualUser Yes, if $ker A =(vec0 )$, then $A$ is injective, and for a square matrix, injectivity is the same as bijectivity.
    $endgroup$
    – Song
    Feb 1 at 17:35












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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









4












$begingroup$

In fact, the conclusion should be that $A+A^T$ is not invertible. Note that the range of $A$ is spanned by
$$
langle Ax_1,Ax_2,ldots , Ax_{10}rangle =langle Ax_7,Ax_8,Ax_{9}, Ax_{10}rangle,
$$ which implies that the rank of $A$ is at most $4$. By the rank theorem, $A^T$ has the same rank as $A$. This implies that $A+A^T$ has rank at most $text{rk}(A)+text{rk}(A^T)le 8<10$, which says that $A+A^T$ is not invertible.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    that's a nice proof!
    $endgroup$
    – Surb
    Feb 1 at 17:23










  • $begingroup$
    @Surb Thank you :-)
    $endgroup$
    – Song
    Feb 1 at 17:24






  • 1




    $begingroup$
    Made all the nicer by the fact its strategy also makes obvious why the original problem equated seven vectors; with just six, this proof wouldn't have worked.
    $endgroup$
    – J.G.
    Feb 1 at 17:28










  • $begingroup$
    So if $ker A$ has only $vec{0}$ then $A$ is irvertible?
    $endgroup$
    – VirtualUser
    Feb 1 at 17:29






  • 1




    $begingroup$
    @VirtualUser Yes, if $ker A =(vec0 )$, then $A$ is injective, and for a square matrix, injectivity is the same as bijectivity.
    $endgroup$
    – Song
    Feb 1 at 17:35
















4












$begingroup$

In fact, the conclusion should be that $A+A^T$ is not invertible. Note that the range of $A$ is spanned by
$$
langle Ax_1,Ax_2,ldots , Ax_{10}rangle =langle Ax_7,Ax_8,Ax_{9}, Ax_{10}rangle,
$$ which implies that the rank of $A$ is at most $4$. By the rank theorem, $A^T$ has the same rank as $A$. This implies that $A+A^T$ has rank at most $text{rk}(A)+text{rk}(A^T)le 8<10$, which says that $A+A^T$ is not invertible.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    that's a nice proof!
    $endgroup$
    – Surb
    Feb 1 at 17:23










  • $begingroup$
    @Surb Thank you :-)
    $endgroup$
    – Song
    Feb 1 at 17:24






  • 1




    $begingroup$
    Made all the nicer by the fact its strategy also makes obvious why the original problem equated seven vectors; with just six, this proof wouldn't have worked.
    $endgroup$
    – J.G.
    Feb 1 at 17:28










  • $begingroup$
    So if $ker A$ has only $vec{0}$ then $A$ is irvertible?
    $endgroup$
    – VirtualUser
    Feb 1 at 17:29






  • 1




    $begingroup$
    @VirtualUser Yes, if $ker A =(vec0 )$, then $A$ is injective, and for a square matrix, injectivity is the same as bijectivity.
    $endgroup$
    – Song
    Feb 1 at 17:35














4












4








4





$begingroup$

In fact, the conclusion should be that $A+A^T$ is not invertible. Note that the range of $A$ is spanned by
$$
langle Ax_1,Ax_2,ldots , Ax_{10}rangle =langle Ax_7,Ax_8,Ax_{9}, Ax_{10}rangle,
$$ which implies that the rank of $A$ is at most $4$. By the rank theorem, $A^T$ has the same rank as $A$. This implies that $A+A^T$ has rank at most $text{rk}(A)+text{rk}(A^T)le 8<10$, which says that $A+A^T$ is not invertible.






share|cite|improve this answer









$endgroup$



In fact, the conclusion should be that $A+A^T$ is not invertible. Note that the range of $A$ is spanned by
$$
langle Ax_1,Ax_2,ldots , Ax_{10}rangle =langle Ax_7,Ax_8,Ax_{9}, Ax_{10}rangle,
$$ which implies that the rank of $A$ is at most $4$. By the rank theorem, $A^T$ has the same rank as $A$. This implies that $A+A^T$ has rank at most $text{rk}(A)+text{rk}(A^T)le 8<10$, which says that $A+A^T$ is not invertible.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Feb 1 at 17:21









SongSong

18.6k21651




18.6k21651












  • $begingroup$
    that's a nice proof!
    $endgroup$
    – Surb
    Feb 1 at 17:23










  • $begingroup$
    @Surb Thank you :-)
    $endgroup$
    – Song
    Feb 1 at 17:24






  • 1




    $begingroup$
    Made all the nicer by the fact its strategy also makes obvious why the original problem equated seven vectors; with just six, this proof wouldn't have worked.
    $endgroup$
    – J.G.
    Feb 1 at 17:28










  • $begingroup$
    So if $ker A$ has only $vec{0}$ then $A$ is irvertible?
    $endgroup$
    – VirtualUser
    Feb 1 at 17:29






  • 1




    $begingroup$
    @VirtualUser Yes, if $ker A =(vec0 )$, then $A$ is injective, and for a square matrix, injectivity is the same as bijectivity.
    $endgroup$
    – Song
    Feb 1 at 17:35


















  • $begingroup$
    that's a nice proof!
    $endgroup$
    – Surb
    Feb 1 at 17:23










  • $begingroup$
    @Surb Thank you :-)
    $endgroup$
    – Song
    Feb 1 at 17:24






  • 1




    $begingroup$
    Made all the nicer by the fact its strategy also makes obvious why the original problem equated seven vectors; with just six, this proof wouldn't have worked.
    $endgroup$
    – J.G.
    Feb 1 at 17:28










  • $begingroup$
    So if $ker A$ has only $vec{0}$ then $A$ is irvertible?
    $endgroup$
    – VirtualUser
    Feb 1 at 17:29






  • 1




    $begingroup$
    @VirtualUser Yes, if $ker A =(vec0 )$, then $A$ is injective, and for a square matrix, injectivity is the same as bijectivity.
    $endgroup$
    – Song
    Feb 1 at 17:35
















$begingroup$
that's a nice proof!
$endgroup$
– Surb
Feb 1 at 17:23




$begingroup$
that's a nice proof!
$endgroup$
– Surb
Feb 1 at 17:23












$begingroup$
@Surb Thank you :-)
$endgroup$
– Song
Feb 1 at 17:24




$begingroup$
@Surb Thank you :-)
$endgroup$
– Song
Feb 1 at 17:24




1




1




$begingroup$
Made all the nicer by the fact its strategy also makes obvious why the original problem equated seven vectors; with just six, this proof wouldn't have worked.
$endgroup$
– J.G.
Feb 1 at 17:28




$begingroup$
Made all the nicer by the fact its strategy also makes obvious why the original problem equated seven vectors; with just six, this proof wouldn't have worked.
$endgroup$
– J.G.
Feb 1 at 17:28












$begingroup$
So if $ker A$ has only $vec{0}$ then $A$ is irvertible?
$endgroup$
– VirtualUser
Feb 1 at 17:29




$begingroup$
So if $ker A$ has only $vec{0}$ then $A$ is irvertible?
$endgroup$
– VirtualUser
Feb 1 at 17:29




1




1




$begingroup$
@VirtualUser Yes, if $ker A =(vec0 )$, then $A$ is injective, and for a square matrix, injectivity is the same as bijectivity.
$endgroup$
– Song
Feb 1 at 17:35




$begingroup$
@VirtualUser Yes, if $ker A =(vec0 )$, then $A$ is injective, and for a square matrix, injectivity is the same as bijectivity.
$endgroup$
– Song
Feb 1 at 17:35


















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